Question

an engineering company has four openings and the applicant pool consists of five database administrators and eight network engineers. if the hiring is done without regard for the specific qualifications of the applicants, find the probability that the four hired will be
(a) all network engineers. round your answer to five decimal places.
the probability that the four people hired will be all network engineers is ​.

Explanation:

Step1: Calculate the total number of ways to choose 4 people from the applicant pool

The total number of applicants is \(5 + 8=13\). The number of ways to choose 4 people from 13 is given by the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 13\) and \(r = 4\).

$$C(13,4)=\frac{13!}{4!(13 - 4)!}=\frac{13\times12\times11\times10}{4\times3\times2\times1}=715$$

Step2: Calculate the number of ways to choose 4 network engineers

The number of network engineers is 8. The number of ways to choose 4 network engineers from 8 is \(C(8,4)\)

$$C(8,4)=\frac{8!}{4!(8 - 4)!}=\frac{8\times7\times6\times5}{4\times3\times2\times1}=70$$

Step3: Calculate the probability

The probability \(P\) that the four people hired are all network engineers is the number of favorable outcomes (choosing 4 network engineers) divided by the number of total outcomes (choosing 4 people from 13)

$$P=\frac{C(8,4)}{C(13,4)}=\frac{70}{715}\approx0.098$$

Answer:

\(0.098\)